You have found the following ages (in years) of all 5 seals at your local zoo: $ 24,\enspace 1,\enspace 7,\enspace 6,\enspace 23$ What is the average age of the seals at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 seals at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{24 + 1 + 7 + 6 + 23}{{5}} = {12.2\text{ years old}} $ Find the squared deviations from the mean for each seal. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $24$ years $11.8$ years $139.24$ years $^2$ $1$ year $-11.2$ years $125.44$ years $^2$ $7$ years $-5.2$ years $27.04$ years $^2$ $6$ years $-6.2$ years $38.44$ years $^2$ $23$ years $10.8$ years $116.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{139.24} + {125.44} + {27.04} + {38.44} + {116.64}} {{5}} $ $ {\sigma^2} = \dfrac{{446.8}}{{5}} = {89.36\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{89.36\text{ years}^2}} = {9.5\text{ years}} $ The average seal at the zoo is 12.2 years old. There is a standard deviation of 9.5 years.